dy/dx - y = x. how do i solve for y(x)? please help.?

Answer 1

Solve first
y' - y = 0
y = A*e^x

Then solve the particular integral
y' - y = x
y = ax + b
y' = a
y' - y = (a-b) - ax = x
a = -1, b = -1

Final solution:
y = Ae^x - (x+1)

Answer 2

This is a non-homogeneous linear first order differential equation. It is solved in two steps; first set the right side to zero to get the general solution:

dy/dx - y = 0

You can verify that this gives y = k*e^x as the solution. To this add a particular solution derived from the right-hand expression. In this case, you use a polynomial in x: ax^2 + bx + c as the particular solution, and add it to the general solution above. Put this result

y = k*e^x + a*x^2 + b*x + c

into the original equation

dy/dx = ke^x + 2*a*x + b

dy/dx - y = ke^x + 2*a*x + b - k*e^x - a*x^2 - b*x - c

= 2*a*x + b - a*x^2 - b*x - c

and this must = the right side, or x, so

2*a*x + b - a*x^2 - b*x - c = x

Equate coefficients of like powers of x to get

a = 0, b - c = 0 and (2*a - b) = 1

b = -1 and c = b = -1

The solution is then

y = k*e^x - x - 1

Answer 3

antidifferentiation..

-y=x dx
(-y^2)/2 = (x^2)/2 + c (c= the y-intercept)
-y^2 = x^2
-y= (sqr root) x
therefore..

y(x) = - (sqr root) x

Answer 4

dy/dx - y = x
x = y/x - y
x = y/x - xy/x
1 = y - xy
y = xy + 1
x = (y - 1)/y

Answer 5

[dy/dx-y]/y=y[x]